Common-Ion Effect同离子效应

Big Idea · Scale, Proportion, and Quantity (SPQ)AP Science Practice 6 · ArgumentationAP link · Unit 7~40 min

Learning goals

  • Use Le Châtelier's principle and the comparison of Q with Ksp to explain why a common ion lowers the solubility of a slightly soluble salt
  • Calculate the molar solubility of a slightly soluble salt in a solution that already contains a common ion
  • Explain that the common-ion effect changes the solubility but not Ksp

Before you start

  • Introduction to Le Châtelier's principle (U7.9)
  • Introduction to solubility equilibria (Ksp) (U7.11)

Less dissolves when one of the ions is already there

The molar solubility of AgCl in pure water is 1.3 × 10⁻⁵ M. If the solution already contains Cl⁻ (an NaCl solution, say), much less AgCl dissolves. This is the common-ion effect同离子效应.

Two ways of saying the same thing:

  • Le Châtelier’s principle: in AgCl(s) ⇌ Ag⁺ + Cl⁻, the product Cl⁻ is already at a high concentration, so the equilibrium shifts toward the solid.
  • Q and Ksp: Ksp = [Ag⁺][Cl⁻] is fixed. The higher [Cl⁻], the lower the [Ag⁺] that can exist at saturation, so less AgCl dissolves (the amount dissolved equals the Ag⁺ it produces).

The common-ion effect changes the solubility; Ksp stays the same (at constant temperature).

Calculating: approximate when the common ion is much more concentrated than s

If the solution already contains a common ion at concentration c, then after s M of salt dissolves the common ion is at c + n·s. When s is small, c + n·s ≈ c, which makes the calculation much simpler. Check afterward that s really is much smaller than c.

Drag the concentration of the common ion and watch the solubility change (both axes are logarithmic):

Try it: the common-ion effect

Example 1: AgCl in NaCl solution

Find the molar solubility at 25 °C of AgCl (Ksp = 1.6 × 10⁻¹⁰) in 0.010 M NaCl, and compare it with pure water.

Step-by-step solution

  1. Identify the common ion. NaCl dissociates completely, so the solution already has [Cl⁻] = 0.010 M; Cl⁻ is the common ion.

  2. Set up an ICE table. Let s M of AgCl dissolve (concentrations in M):

    AgCl(s)Ag⁺Cl⁻
    I—00.010
    C−s+s+s
    E—s0.010 + s

  3. Substitute into Ksp.

    Ksp=s (0.010+s)=1.6×10−10K_{sp} = s\,(0.010 + s) = 1.6 \times 10^{-10}

  4. Approximate. Even in pure water AgCl dissolves only to about 10⁻⁵ M, and less with a common ion, so s ≪ 0.010 and 0.010 + s ≈ 0.010:

    0.010 s=1.6×10−10s=1.6×10−100.010=1.6×10−8 M\begin{aligned} 0.010\,s &= 1.6 \times 10^{-10} \\ s &= \frac{1.6 \times 10^{-10}}{0.010} \\ &= \boxed{1.6 \times 10^{-8}\ \text{M}} \end{aligned}

  5. Check the approximation. 1.6 × 10⁻⁸ ≪ 0.010 (only 10⁻⁶ of it), so the approximation is reasonable.

  6. Compare with pure water. With no common ion, [Ag⁺] = [Cl⁻] = s:

    s2=1.6×10−10s=1.6×10−10=1.3×10−5 M\begin{aligned} s^2 &= 1.6 \times 10^{-10} \\ s &= \sqrt{1.6 \times 10^{-10}} \\ &= 1.3 \times 10^{-5}\ \text{M} \end{aligned} 1.3×10−51.6×10−8≈800\frac{1.3 \times 10^{-5}}{1.6 \times 10^{-8}} \approx 800

    In 0.010 M NaCl, AgCl is only about 1/800 as soluble as in pure water.

Example 2: Common cation or common anion?

Compare the molar solubility of PbI₂ (Ksp = 1.4 × 10⁻⁸) in 0.10 M KI and in 0.10 M Pb(NO₃)₂.

Step-by-step solution

  1. Write the Ksp expression.

    PbIX2(s)⇌PbX2++2 IX−Ksp=[PbX2+][IX−]2\begin{gathered} \ce{PbI2(s) <=> Pb^2+ + 2I-} \\ K_{sp} = [\ce{Pb^2+}][\ce{I-}]^2 \end{gathered}

  2. In 0.10 M KI, the common ion is I⁻. [Pb²⁺] = s and [I⁻] = 0.10 + 2s ≈ 0.10:

    s×(0.10)2=1.4×10−8s=1.4×10−80.010=1.4×10−6 M\begin{aligned} s \times (0.10)^2 &= 1.4 \times 10^{-8} \\ s &= \frac{1.4 \times 10^{-8}}{0.010} \\ &= \boxed{1.4 \times 10^{-6}\ \text{M}} \end{aligned}

  3. In 0.10 M Pb(NO₃)₂, the common ion is Pb²⁺. [Pb²⁺] = 0.10 + s ≈ 0.10 and [I⁻] = 2s:

    0.10×(2s)2=1.4×10−80.40 s2=1.4×10−8s2=3.5×10−8s=1.9×10−4 M\begin{aligned} 0.10 \times (2s)^2 &= 1.4 \times 10^{-8} \\ 0.40\,s^2 &= 1.4 \times 10^{-8} \\ s^2 &= 3.5 \times 10^{-8} \\ s &= \boxed{1.9 \times 10^{-4}\ \text{M}} \end{aligned}

  4. Check the approximations. In both cases s and 2s are far smaller than 0.10, so the approximations hold.

  5. Compare and explain. In pure water s = 1.5 × 10⁻³ M (U7.11, Example 1). Divide it by the solubility in each solution:

    swatersKI=1.5×10−31.4×10−6≈1000swatersPb(NOX3)X2=1.5×10−31.9×10−4≈8\begin{aligned} \frac{s_\text{water}}{s_{\ce{KI}}} &= \frac{1.5 \times 10^{-3}}{1.4 \times 10^{-6}} \approx 1000 \\ \frac{s_\text{water}}{s_{\ce{Pb(NO3)2}}} &= \frac{1.5 \times 10^{-3}}{1.9 \times 10^{-4}} \approx 8 \end{aligned}

    At the same 0.10 M, the common I⁻ lowers the solubility far more, because I⁻ is squared in Ksp: with [I⁻] fixed at 0.10, (0.10)² = 0.010 divides straight into Ksp.

Example 3: Ca₃(PO₄)₂ in Na₃PO₄ solution

Find the molar solubility of Ca₃(PO₄)₂ (Ksp = 1.3 × 10⁻³², the value given in the problem) in 0.20 M Na₃PO₄, and compare it with its molar solubility in pure water.

Step-by-step solution

  1. Write the equilibrium and the Ksp expression. For Ca₃(PO₄)₂(s) ⇌ 3Ca²⁺(aq) + 2PO₄³⁻(aq), the solid does not appear in the expression:

    Ksp=[CaX2+]3[POX4X3−]2K_{sp} = [\ce{Ca^2+}]^3[\ce{PO4^3-}]^2

  2. Identify the common ion and its starting concentration. Na₃PO₄ is a soluble strong electrolyte and dissociates completely:

    NaX3POX4→3 NaX++POX4X3−\ce{Na3PO4 -> 3Na+ + PO4^3-}

    Each Na₃PO₄ gives only one PO₄³⁻, so the starting [PO₄³⁻] is 0.20 M (0.60 M is [Na⁺]; don’t substitute that). Na⁺ is not part of this equilibrium and can be ignored.

  3. Set up an ICE table. Let s mol of Ca₃(PO₄)₂ dissolve per liter; each formula unit gives 3 Ca²⁺ and 2 PO₄³⁻ (concentrations in M):

    Ca₃(PO₄)₂(s)3Ca²⁺2PO₄³⁻
    I—00.20
    C−s+3s+2s
    E—3s0.20 + 2s

  4. Substitute into Ksp. The coefficients 3 and 2 each appear twice: once inside the concentrations (3s, 2s) and once as exponents:

    (3s)3 (0.20+2s)2=1.3×10−32(3s)^3\,(0.20 + 2s)^2 = 1.3 \times 10^{-32}

  5. Approximate. Even in pure water the solubility is only about 10⁻⁷ M (step 8), and the common ion makes it smaller still, so 0.20 + 2s ≈ 0.20:

    (3s)3×(0.20)2=1.3×10−3227s3×0.040=1.3×10−321.08 s3=1.3×10−32\begin{aligned} (3s)^3 \times (0.20)^2 &= 1.3 \times 10^{-32} \\ 27s^3 \times 0.040 &= 1.3 \times 10^{-32} \\ 1.08\,s^3 &= 1.3 \times 10^{-32} \end{aligned}

  6. Solve for s. Before taking the cube root, make the power of 10 a multiple of 3:

    s3=1.3×10−321.08=1.2×10−32=12×10−33s=123×10−11=2.29×10−11≈2.3×10−11 M\begin{aligned} s^3 &= \frac{1.3 \times 10^{-32}}{1.08} = 1.2 \times 10^{-32} \\ &= 12 \times 10^{-33} \\ s &= \sqrt[3]{12} \times 10^{-11} = 2.29 \times 10^{-11} \\ &\approx \boxed{2.3 \times 10^{-11}\ \text{M}} \end{aligned}

    Then [Ca²⁺] = 3s = 6.9 × 10⁻¹¹ M.

  7. Check the approximation.

    2s0.20=4.6×10−110.20=2.3×10−10\frac{2s}{0.20} = \frac{4.6 \times 10^{-11}}{0.20} = 2.3 \times 10^{-10}

    2s is only about 2 × 10⁻¹⁰ of 0.20, so the approximation is excellent.

  8. Compare with pure water. With no common ion, [Ca²⁺] = 3s and [PO₄³⁻] = 2s:

    Ksp=(3s)3(2s)2=27×4×s5=108 s5\begin{aligned} K_{sp} &= (3s)^3(2s)^2 \\ &= 27 \times 4 \times s^5 = 108\,s^5 \end{aligned} s5=1.3×10−32108=12×10−35s=125×10−7=1.6×10−7\begin{aligned} s^5 &= \frac{1.3 \times 10^{-32}}{108} = 12 \times 10^{-35} \\ s &= \sqrt[5]{12} \times 10^{-7} = 1.6 \times 10^{-7} \end{aligned} 1.64×10−72.29×10−11≈7×103\frac{1.64 \times 10^{-7}}{2.29 \times 10^{-11}} \approx 7 \times 10^{3}

    In 0.20 M Na₃PO₄, Ca₃(PO₄)₂ is only about 1/7000 as soluble as in pure water.

Note: PO₄³⁻ is a base and reacts with water to form HPO₄²⁻ (Unit 8), so in a real solution [PO₄³⁻] is somewhat less than 0.20 M. AP common-ion calculations usually ignore this.

Practice: each substance is added to a saturated solution (volume about the same). How does the solubility of the slightly soluble salt change?
  1. NaCl added to AgCl
  2. AgNO₃ added to AgCl
  3. KNO₃ added to AgCl
  4. Na₂CO₃ added to CaCO₃
  5. NaOH added to Mg(OH)₂
  6. HCl added to Mg(OH)₂
  7. Pb(NO₃)₂ added to PbI₂
  8. NaNO₃ added to PbSO₄

Common mistakes

Mistake 1: “Adding a common ion makes Ksp smaller.” Ksp changes only with temperature. What decreases is the solubility: with more of one ion present, less of the other can exist.

Mistake 2: “Adding any salt lowers the solubility of a slightly soluble salt.” Only a salt that shares an ion causes the common-ion effect. Adding something that uses up one of the ions (such as an acid using up OH⁻) actually raises the solubility.

Mistake 3: “For PbI₂ in 0.10 M Pb(NO₃)₂, [I⁻] = s.” [I⁻] is 2s, so (2s)² goes into Ksp.

Quick check

Ksp for BaF₂ is 2.4 × 10⁻⁵ (OpenStax). Its molar solubility in 0.10 M NaF is about:

Solid NaCl is added to a saturated AgCl solution. Which statement is correct?

In which solution is CaCO₃ least soluble?

Ag⁺ in a solution is precipitated as AgCl using NaCl. To remove the Ag⁺ more completely, you should:

Which statement explains the common-ion effect?

The ratio of the molar solubility of AgCl in 0.0010 M NaCl to that in 0.10 M NaCl is about:

AP-style free response

AP-style free response

Hard water contains Ca²⁺. One way to soften it is to add Na₂CO₃ so that the Ca²⁺ precipitates as CaCO₃. At 25 °C, Ksp for CaCO₃ is 8.7 × 10⁻⁹ (OpenStax).

(a) Calculate the molar solubility of CaCO₃ in pure water.

(b) Calculate the molar solubility of CaCO₃ in 0.010 M Na₂CO₃, and check your approximation.

(c) A sample of hard water has [Ca²⁺] = 1.0 × 10⁻³ M. Na₂CO₃ is added so that the total CO₃²⁻ concentration is 0.010 M (ignore the change in volume). After precipitation is complete, how much Ca²⁺ remains, and what percentage of the Ca²⁺ has been removed?

(d) Use Le Châtelier’s principle to explain why excess CO₃²⁻ precipitates the Ca²⁺ more completely.

(e) A student thinks that adding enough Na₂CO₃ will remove the Ca²⁺ completely (to zero concentration). Identify the error in this idea.

Model answer and scoring

(a) Ksp = s², so s = √(8.7 × 10⁻⁹) = 9.3 × 10⁻⁵ M.

(b) [CO₃²⁻] ≈ 0.010, so s = 8.7 × 10⁻⁹ ÷ 0.010 = 8.7 × 10⁻⁷ M. Since 8.7 × 10⁻⁷ ≪ 0.010, the approximation is reasonable.

(c) Precipitation uses about 1.0 × 10⁻³ M of CO₃²⁻, leaving [CO₃²⁻] ≈ 0.009 M; then [Ca²⁺] = 8.7 × 10⁻⁹ ÷ 0.009 ≈ 9.7 × 10⁻⁷ M. The fraction removed is (1.0 × 10⁻³ − 9.7 × 10⁻⁷) ÷ 1.0 × 10⁻³ ≈ 99.9%.

(d) In CaCO₃(s) ⇌ Ca²⁺ + CO₃²⁻, raising [CO₃²⁻] shifts the equilibrium toward solid CaCO₃, so less Ca²⁺ remains in solution.

(e) At equilibrium [Ca²⁺] = Ksp ÷ [CO₃²⁻]. As long as [CO₃²⁻] is finite, [Ca²⁺] can’t be zero; it can only be made very small.

Scoring: (a) s; (b) approximation and check; (c) remaining CO₃²⁻, remaining Ca²⁺, and the percentage; (d) the shift; (e) Ksp keeps the ion concentration from reaching zero.

Summary

  • When one of its ions is already present, a slightly soluble salt dissolves less (the common-ion effect); Ksp is unchanged.
  • To calculate, substitute the concentration of the common ion into Ksp; when s is small, approximate c + n·s ≈ c, then check.
  • A common ion that appears squared in Ksp lowers the solubility more strongly.

AP tip: To explain the common-ion effect, spell out “the concentration of ion … increases → Q > Ksp (or the equilibrium shifts left) → the solubility decreases,” and state that Ksp doesn’t change. When calculating, write expressions for both ion concentrations (c + n·s and m·s) before approximating.

Practice this chapter: Unit 7 · Equilibrium · 12 questionsPractice this topic: U7.12 · Common-Ion Effect · 6 questions

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